What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of $\bar{x}=\$ 6.88$ per 100 pounds of watermelon. Assume that $\sigma$ is known to be $\$ 1.94$ per 100 pounds.
(a) Find a $90 \%$ confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (Round your answers to two decimal places.)
lower limit
upper limit
margin of error
(b) Find the sample size necessary for a $90 \%$ confidence level with maximal error of estimate $E=0.37$ for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)
farming regions
(c) A farm brings 15 tons of watermelon to market. Find a $90 \%$ confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)
lower limit
upper limit
margin of error

Step 1 ：Calculate the 90% confidence interval for the population mean price using the formula: \(CI = \bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation, and \(n\) is the sample size.

Step 2 ：For a 90% confidence level, the z-score is approximately 1.645. So, the confidence interval is: \(CI = 6.88 \pm 1.645*\frac{1.94}{\sqrt{45}}\)

Step 3 ：Calculate the above expression to get: \(CI = 6.88 \pm 0.50\)

Step 4 ：The lower limit is \(6.88 - 0.50 = 6.38\) and the upper limit is \(6.88 + 0.50 = 7.38\). The margin of error is \(0.50\). \(\boxed{CI = [6.38, 7.38]}\)

Step 5 ：Calculate the sample size necessary for a 90% confidence level with maximal error of estimate E=0.37 using the formula: \(n = (\frac{z*\sigma}{E})^2\)

Step 6 ：Substitute the given values to get: \(n = (\frac{1.645*1.94}{0.37})^2\)

Step 7 ：Calculate the above expression to get: \(n \approx 69.4\)

Step 8 ：Since we can't have a fraction of a farming region, we round up to the nearest whole number, so the required number of farming regions is 70. \(\boxed{n = 70}\)

Step 9 ：Calculate the 90% confidence interval for the population mean cash value of this crop. First, convert the weight of the watermelon from tons to 100 pounds units, since the price is given per 100 pounds. 1 ton is 2000 pounds, so 15 tons is 30000 pounds, which is 300 units of 100 pounds.

Step 10 ：Then, use the formula for the confidence interval: \(CI = \bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)

Step 11 ：Substitute the given values to get: \(CI = 6.88*300 \pm 1.645*\frac{1.94*300}{\sqrt{45}}\)

Step 12 ：Calculate the above expression to get: \(CI = 2064 \pm 150\)

Step 13 ：The lower limit is \(2064 - 150 = 1914\) and the upper limit is \(2064 + 150 = 2214\). The margin of error is \(150\). \(\boxed{CI = [1914, 2214]}\)